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a(2n) = n*(n+1)/2, a(2n+1) = n!.
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%I #10 Sep 03 2016 17:05:45

%S 0,1,1,1,3,2,6,6,10,24,15,120,21,720,28,5040,36,40320,45,362880,55,

%T 3628800,66,39916800,78,479001600,91,6227020800,105,87178291200,120,

%U 1307674368000,136,20922789888000,153,355687428096000,171

%N a(2n) = n*(n+1)/2, a(2n+1) = n!.

%C a(n) = n + a(n-1), a(n) = n*(n+1)/2, with a(0) = 0 and b(n) = n*a(n-1), a(n) = n!, with b(0) = 1, then you can form a function F such that: F(n!) = n*(n+1)/2.

%H G. C. Greubel, <a href="/A154028/b154028.txt">Table of n, a(n) for n = 0..900</a>

%t Flatten[Table[{n*(n + 1)/2, n!}, {n, 0, 20}]] (* Produces terms *)

%t (*addition*)

%t a[0] = 0; a[n_] := a[n] = n + a[n - 1];

%t Table[a[n] - n*(n + 1)/2, {n, 0, 20}] (* Demonstrates that a[n] = n*(n+1)/2 *)

%t (*multiplication*)

%t b[0] = 1; b[n_] := b[n] = n*b[n - 1];

%t Table[b[n] - n!, {n, 0, 20}] (* Demonstrates that b[n] = n! *)

%o (PARI) a(n)=if(n%2,(n\2)!,n*(n+2)/8) \\ _Charles R Greathouse IV_, Sep 01 2016

%K nonn,easy

%O 0,5

%A _Roger L. Bagula_, Jan 04 2009